x is divided by y but not partial order

Question

One of the problems in my discrete math course says: Let $x,y \in \mathbb N-\{0\}$. $(x,y) \in D$ iff $x$ divides $y$ or $y$ divides $x$. The problems asks whether this is a partial order or not and the answer is that it's not. The only way this would be true is if $x \neq y$. Are we supposed to assume that $x \neq y$? Because if $x=y$ then this should be a partial order because it's reflexive, antisymmetric and transitive.

Answer 1

The definition of the relation $D$ is made for any pair of positive integers. You must not assume that $x=y$ or that $x\neq y$.

$D$ is not a partial order because it is not transitive. A relation is not transitive when there is at least one conterexample. That is, to show that $D$ is not transitive, you only need to find two pairs $(x,y)$ and $(y,z)$ in $D$ such that $(x,z)$ is not in $D$.

$(3,6)$ and $(6,2)$ are in $D$, but $(3,2)$ is not.

Answer 2

It is clearly symmetric, the definition itself is symmetric.

Answer 3

Are you sure it's transitive? Think about the sequence $2, 6, 3$ . . .

Also, it's not antisymmetric: $(2, 6)$ and $(6,2)$ are both in $D$.